Problem: Find the polynomial $p(x),$ with real coefficients, such that
\[p(x^3) - p(x^3 - 2) = [p(x)]^2 + 12\]for all real numbers $x.$
Explanation: Let
\[p(x) = a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0,\]where $a_n \neq 0.$  Then
\begin{align*}
p(x^3) - p(x^3 - 2) &= a_n x^{3n} + a_{n - 1} x^{3n - 3} + \dotsb - a_n (x^3 - 2)^n - a_{n - 1} (x^3 - 2)^{n - 1} - \dotsb \\
&= a_n x^{3n} + a_{n - 1} x^{3n - 3} + \dotsb - a_n x^{3n} - 2na_n x^{3n - 3} - \dotsb - a_{n - 1} x^{3n - 3} - \dotsb \\
&= 2n a_n x^{3n - 3} + \dotsb.
\end{align*}Thus, the degree of $p(x^3) - p(x^3 - 2)$ is $3n - 3.$

The degree of $[p(x)^2] + 12$ is $2n,$ so $3n - 3 = 2n,$ which means $n = 3.$

Let $p(x) = ax^3 + bx^2 + cx + d.$  Then
\begin{align*}
p(x^3) - p(x^3 - 2) &= ax^9 + bx^6 + cx^3 + d - (a(x^3 - 2)^3 + b(x^3 - 2)^2 + c(x^3 - 2) + d) \\
&= 6ax^6 + (-12a + 4b) x^3 + 8a - 4b + 2c,
\end{align*}and
\[[p(x)]^2 + 12 = a^2 x^6 + 2abx^5 + (2ac + b^2) x^4 + (2ad + 2bc) x^3 + (2bd + c^2) x^2 + 2cdx + d^2 + 12.\]Comparing coefficients, we get
\begin{align*}
a^2 &= 6a, \\
2ab &= 0, \\
2ac + b^2 &= 0, \\
2ad + 2bc &= -12a + 4b, \\
2bd + c^2 &= 0, \\
2cd &= 0, \\
d^2 + 12 &= 8a - 4b + 2c.
\end{align*}From the equation $a^2 = 6a,$ $a = 0$ or $a = 6.$  But since $a$ is a leading coefficient, $a$ cannot be 0, so $a = 6.$

From the equation $2ab = 0,$ $b = 0.$

Then the equation $2ac + b^2 = 0$ becomes $12c = 0,$ so $c = 0.$

Then the equation $2ad + 2bc = -12a + 4b$ becomes $12d = -72,$ so $d = -6.$  Note that $(a,b,c,d) = (6,0,0,-6)$ satisfies all the equations.

Therefore, $p(x) = \boxed{6x^3 - 6}.$